September 15

Briefly, in March 2013 (study the history of the 1 article) there was a period where someone wrote that 1 to the infinity power is e. Is there any logic in this statement?? Georgia guy (talk) 21:46, 15 September 2021 (UTC)[reply]

One of the classic ways of defining e is

${\displaystyle e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}}$

which is an example of the indeterminate form 1^∞. So you can sort of see where it comes from. As to whether this is a better choice of a value to assign to 1^∞ than any other, I would say that it is not. --Trovatore (talk) 21:49, 15 September 2021 (UTC)[reply]

In general:

${\displaystyle e^{x}=\lim _{n\to \infty }\left(1+{\frac {x}{n}}\right)^{n}.}$

So by changing the value of x you can get any positive value for 1^∞. If you allow complex values for x then you can get any non-zero value. You could tweak the limit in other ways to get 1^∞ = 0 or ∞. I think it's best to leave 1^∞, along with 0/0, ∞/∞, ∞ − ∞, etc. undefined except as indeterminate forms with no assigned value. --RDBury (talk) 22:20, 15 September 2021 (UTC)[reply]

PS. "Briefly" here means for approximately 8 min. on March 11. There is also a blurb in the talk page Talk:1/Archive 1#1^∞ is e. --RDBury (talk) 22:50, 15 September 2021 (UTC)[reply]

It is indeed best left alone. One might choose to define it that way, but that makes a lot of algebra break down. You need to introduce all kinds of exceptions all over the place, or else you get things like

${\displaystyle e^{2}=(1^{\infty })^{2}=1^{2\times \infty }=(1^{2})^{\infty }=1^{\infty }=e.}$

 --Lambiam 10:25, 18 September 2021 (UTC)[reply]

"I would quarrel with mathematics and say that the sum of many zeros is a dangerous number" (Stanislaw Jerzy Lec). And ${\displaystyle 1^{\infty }}$ is the exponential of it. pma 13:53, 21 September 2021 (UTC)[reply]

The sum of any number of zeros is zero. Georgia guy (talk) 14:40, 21 September 2021 (UTC)[reply]

Corollary. Zero is a dangerous number.  --Lambiam 14:48, 21 September 2021 (UTC)[reply]

@Georgia guy: AFAIK the 'zero' in the Lec's aphorism may be a person with little knowledge (possibly also little skills) compared to their social position. --CiaPan (talk) 19:19, 21 September 2021 (UTC)[reply]

See 1 + 2 + 3 + 4 + ⋯ for some other strange formulas, but I don't remember 1^∞=e having anything like that. 67.164.113.165 (talk) 05:31, 22 September 2021 (UTC)[reply]

September 17

2-Euro coins weigh 8.5 gram. 1-Euro coins weigh 7.5 gram. How many of each would you need to have exactly 1 kg?

I solved the problem by trial and error, limiting the max nr. of coins to 1000/8.5=117. The results are: 10, 25, 40, 55, 70, 85, 100, 115x 2-Euro coins and the 1-Euro complement.

for i in range(1,117): 
....     if (1000 - 8.5*i) % 7.5 == 0: 
....         print(f'{i}x 2 Euro coins and {int((1000-i*8.5)/7.5)}x 1 Euro coins.') 

Is there a more direct way of solving this and similar kinds of problems? --Bumptump (talk) 23:28, 17 September 2021 (UTC)[reply]

@Bumptump: Convert it to a problem with integers: 17x + 15y = 2000. See Diophantine equation#Linear Diophantine equations. PrimeHunter (talk) 23:37, 17 September 2021 (UTC)[reply]

It's not said, but I assume we're not including solutions with negative coins. That makes the question a bit trickier. Variation: What's the largest multiple of .5 grams that can't be weighed out with 1 & 2 Euro coins (also assuming no negative coins)? --RDBury (talk) 06:11, 18 September 2021 (UTC)[reply]

A coin with negative mass will be worth much more than € 2.  --Lambiam 10:11, 18 September 2021 (UTC)[reply]

If we are weighing using a traditional scale (with two plates), it makes totally sense to talk about negative coins. No need to get coins with negative mass. 130 * 8.5 = 1000 + 14 * 7.5. --Bumptump (talk) 16:18, 18 September 2021 (UTC)[reply]

With two coin types we can use the explicit formula given by xy - x - y, here yielding 223/2 as the largest that cannot be made using 8.5 and 7.5; a solution exists for all higher amounts and can be computed explicitly and efficiently with the (extended) Euclidean algorithm. The problem with three or more coin types (and no negative coins) is NP-hard; see coin problem.--Jasper Deng (talk) 22:51, 18 September 2021 (UTC)[reply]

September 19

For a 2n-polygon with side of 1, what fraction of the area is found in the "center rectangle" formed of the points 1, 2 , n+1 & n+2? It looks somewhat easily to figure out "by hand" for the Hexagon and the Octagon, but beyond that, I'm not sure. It certainly goes to zero and n goes to infinity.Naraht (talk) 06:13, 19 September 2021 (UTC)[reply]

2/n catslash (talk) 14:53, 19 September 2021 (UTC)[reply]

Let the side length be ${\displaystyle s}$ and let the distance from the centre of the (regular) polygon to the middle of a side (the apothem) be ${\displaystyle h}$. Then the area of the polygon can be divided into ${\displaystyle 2n}$ triangles, each of area ${\displaystyle {\frac {1}{2}}sh}$. The area of the rectangle is ${\displaystyle 2hs}$, so the fractional area is

${\displaystyle f={\frac {2hs}{2n{\frac {1}{2}}sh}}={\frac {2}{n}}}$

Sanity check: for a square ${\displaystyle f=1}$ and for a circle ${\displaystyle f=0}$. catslash (talk) 15:11, 19 September 2021 (UTC)[reply]

Another way to look at this is to divide the rectangle along it's diagonals. Each of the four triangles formed has the same area, so one triangle is 1/4 the area of the rectangle. But, as you pointed out, 2n of the thicker type triangles can be put together to form the entire polygon, and so each triangle is 1/2n the area of the polygon and the result follows. You could express the area of the rectangle and the polygon in terms of sine and cosine, but your way is much more elegant.

Another variation is to divide the rectangle into 8 congruent triangles like the Union Jack. This time 4n of the triangles can be fit together to form the polygon and rest is similar.

If you divide the rectangle along one diagonal, the you get a triangle inscribed in a regular m-gon, with vertices of the triangle also vertices of the polygon, and the area of the triangle is a rational multiple of the area of the m-gon. What other instances of this are there? It seems there are none with m=5, but with m=6 any triangle inscribed in this way works. Are there any instance with odd m>3? --RDBury (talk) 11:58, 20 September 2021 (UTC)[reply]

Dissection of the rectangle into 8 triangles, and the polygon into 4n triangles, all congruent, surely wins the prize for elegance. Given a picture, a cat could understand it. catslash (talk) 20:43, 20 September 2021 (UTC)[reply]

September 20

On the page of Normal fan on Wikipedia as one of the properties of normal fans it is written: "If F is a face of P of dimension d, then its normal cone C_F has dimension n – d." For this page there is only one reference and I couldn't find anywhere on this reference such an statement. Is it possible to suggest a more precise reference for this property?Mathlover854 (talk) 12:24, 20 September 2021 (UTC)[reply]

September 22